the equation of the perpendicular bisectors of sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0respectively - Maths - Straight Lines

Question

the equation of the perpendicular bisectors of sides AB and AC
of a triangle ABC are x-y+5=0 and x+2y=0respectively if the point
is A (1,-2), find the equation of line BC

Gursheen kaur. · Accepted Answer

Equation of the perpendicular bisector of AB is x-y+5=0.........(i)

Also, Equation of AB (is perpendicular to x-y+5=0) can be written as y +x+λ=0, but passes through A = (1,-2)

⇒-2+1+λ=0

⇒λ=1

∴AB = y+x+1=0..............(ii)

From (i) and (ii) , we get,

The coordinates of D, the middle point of AB

⇒D=(3,2)

⇒x=-3,y=2

Now, D is the middle point of AB,

Let the coordinates of B be (α,β)

Then,

$\frac{α + 1}{2} = - 3 a n d \frac{β - 2}{2} = 2 \Rightarrow α = - 6 - 1 = - 7 a n d β = 4 + 2 = 6 \Rightarrow B = (α, β) = (- 7, 6) ................ (i i i)$

Now, equation of AC,

2x-y-4=0.................(iv)

Solve this equation by x+2y=0, we get, the coordinates of E , which is the middle point of AC,

$E = (\frac{8}{5}, \frac{- 4}{5})$

Let the coordinate of C be (p, q) , then

$\frac{p + 1}{2} = \frac{8}{5} a n d \frac{q - 2}{2} = \frac{4}{5} \Rightarrow 5 p + 5 = 16 a n d 5 q - 10 = 8 \Rightarrow p = \frac{11}{5} a n d q = \frac{2}{5} \Rightarrow C = (p, q) = (\frac{11}{5}, \frac{2}{5})$

Thus, the equation of BC , which is passing through B and C , is given by

y-q= $\frac{β - q}{α - p} (x - p)$

⇒14x+23y=40

the equation of the perpendicular bisectors of sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0respectively. if the point is A (1,-2), find the equation of line BC.